Question

Differentiate x2/3 - y2/3 = 6 with respect to x and evaluate the derivative at (8, 1).

Answer 1

Implicit Differentiation

Step-by-step explanation:

Given:

$x^{\frac{2}{3}} -y^{\frac{2}{3}} = 6$

Recall:

$x^{\frac{2}{3}} -y^{\frac{2}{3}} = 6 \\ \frac{\text{d}}{\text{d}x}(x^{\frac{2}{3}} -y^{\frac{2}{3}}) = \frac{\text{d}}{\text{d}x}(6) \\ \frac{\text{d}}{\text{d}x}(x^{\frac{2}{3}}) -\frac{\text{d}}{\text{d}x}(y^{\frac{2}{3}}) = 0 \\ \frac{2}{3}x^{\frac{2}{3} -1} -\frac{2}{3}y^{\frac{2}{3} -1}\cdot \frac{\text{d}y}{\text{d}x} = 0 \\ \frac{2}{3}x^{-\frac{1}{3}} - \frac{2}{3}y^{-\frac{1}{3}} \cdot \frac{\text{d}y}{\text{d}x} = 0 \\ \frac{2}{3\sqrt[3]{x}} -\frac{2}{3\sqrt[3]{y}}\cdot \frac{\text{d}y}{\text{d}x} = 0 \\ -\frac{2}{3\sqrt[3]{y}}\cdot \frac{\text{d}y}{\text{d}x} = -\frac{2}{3\sqrt[3]{x}} \\ -2 \cdot \frac{\text{d}y}{\text{d}x} = -\frac{6\sqrt[3]{y}}{3\sqrt[3]{x}} \\ \frac{\text{d}y}{\text{d}x} = -\frac{6\sqrt[3]{y}}{-6\sqrt[3]{x}} \\ \frac{\text{d}y}{\text{d}x} = \frac{\sqrt[3]{y}}{\sqrt[3]{x}}$